244 CHAPTER 11. POWER SERIES
Proof: See Definition 3.3.16 for the notion of limsup and liminf. Note
lim supk→∞
∣∣∣ak (x−a)k∣∣∣1/k
= lim supk→∞
|ak|1/k |x−a| .
Then by the root test, the series converges absolutely if
|x−a| lim supk→∞
|ak|1/k < 1
and diverges if|x−a| lim sup
k→∞
|ak|1/k > 1.
Thus define
r ≡
1/ limsupk→∞ |ak|1/k if ∞ > limsupk→∞ |ak|1/k > 0∞ if limsupk→∞ |ak|1/k = 00 if limsupk→∞ |ak|1/k = ∞
Next let λ be as described. Then if |x−a| ≤ λ , then
lim supk→∞
∣∣∣ak (x−a)k∣∣∣1/k
= lim supk→∞
|ak|1/k |x−a| ≤ λ lim supk→∞
|ak|1/k ≤ λ
r< α < 1
It follows that for all k large enough and such x,∣∣∣ak (x−a)k
∣∣∣< αk. Then by the WeierstrassM test, convergence is uniform.
Note that the radius of convergence r is given by
lim supk→∞
|ak|1/k r = 1
Definition 11.1.4 The number in the above theorem is called the radius of con-vergence and the set on which convergence takes place is called the disc of convergence.Since this book only considers functions of one real variable, it will be called the intervalof convergence.
Now the theorem was proved using the root test but often you use the ratio test to findthe interval of convergence. This kind of thing is typical in math so get used to it. Theproof of a theorem does not always yield a way to find the thing the theorem speaks about.The above is an existence theorem. There exists an interval of convergence from the abovetheorem. You find it in specific cases any way that is most convenient.
Example 11.1.5 Find the interval of convergence of the Taylor series ∑∞n=1
xn
n .
Use Corollary 6.3.3.
limn→∞
(|x|n
n
)1/n
= limn→∞
|x|n√
n= |x|
because limn→∞n√
n = 1 and so if |x|< 1 the series converges. The points satisfying |z|= 1require special attention. When x = 1 the series diverges because it reduces to ∑
∞n=1
1n . At